For a sparingly soluble salt of type $\text{BA}_2$, the dissociation equilibrium is: $\text{BA}_2(s) \rightleftharpoons \text{B}^{2+}(aq) + 2\text{A}^-(aq)$

If $S$ is the molar solubility of the compound, then the equilibrium concentrations of the ions are: $[\text{B}^{2+}] = S$ $[\text{A}^-] = 2S$

The solubility product constant ($K_{sp}$) is given by: $K_{sp} = [\text{B}^{2+}][\text{A}^-]^2$ $K_{sp} = (S)(2S)^2 = 4S^3$

Given that $K_{sp} = 4 \times 10^{-12}$: $4S^3 = 4 \times 10^{-12}$ $S^3 = 10^{-12}$

Taking the cube root on both sides: $S = 10^{-4} \text{ mol/L}$

Thus, the solubility of the compound is $10^{-4} \text{ mol/L}$.